An elementary lemma of commutative algebra - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T12:08:39Z http://mathoverflow.net/feeds/question/40713 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40713/an-elementary-lemma-of-commutative-algebra An elementary lemma of commutative algebra TmobiusX 2010-10-01T04:53:52Z 2010-10-01T14:08:02Z <p>Let $R$ be a commutative ring, $M$, $N$ $R$-modules, and $f: M\rightarrow N$ a homomorphism. It is known that $f$ is injective (surjective) if and only if $f_m$ is injective (surjective) for all maximal ideal $m$. But I don't know whether $f$ is split if and only if $f_m$ is split? Maybe it is true for finitely generated modules over noetherian rings, I suspect. </p> http://mathoverflow.net/questions/40713/an-elementary-lemma-of-commutative-algebra/40714#40714 Answer by Emerton for An elementary lemma of commutative algebra Emerton 2010-10-01T05:16:26Z 2010-10-01T05:21:30Z <p>Let $0 \to A \to B \to C \to 0$ be a short exact sequence of $R$-modules, with $R$ Noetherian and $C$ of finite type, giving rise to a class $c \in Ext^1_R(C,A)$. Tensoring with $R_{\mathfrak m}$ for a maximal ideal $\mathfrak m$ gives the short exact sequence $0 \to A_{\mathfrak m} \to B_{\mathfrak m} \to C_{\mathfrak m} \to 0$, which corresponds to the image of $c$ in the $Ext^1_{R_{\mathfrak m}}(C_{\mathfrak m},A_{\mathfrak m}) = R_{\mathfrak m}\otimes_R Ext^1_R(C,A)$. (This isomorphism holds because $R_{\mathfrak m}$ is flat over $R$; to prove it one computes the Ext via a resolution of $C$ by finite rank free $R$-modules, which exists since $R$ is Noetherian and $C$ is of finite type.)</p> <p>Now suppose that these localizations are all split. Then we have an element $c$ in the $R$-module $Ext^1_R(C,A)$ whose image in the localization at each maximal ideal $\mathfrak m$ vanishes. It follows that this element itself vanishes. (Its annihilator is not contained in any maximal ideal.)</p> <p>So the answer seems to be "yes": for Noetherian $R$, if a short exact sequence whose third term is finite type splits locally at every $\mathfrak m$, it splits.</p> <p>Note: I think one can also see this by comparing sheaf Exts with actual Exts for the associated quasi-coherent sheaves on Spec $R$. The point is that there is a spectral sequence involving the cohohomology of the sheaf Exts converging to the actual Exts (in some generality), but on the affine Spec $R$ the higher cohomology of the sheaf Exts vanishes, so that a class in any $Ext^i$ is a global section of the sheaf Ext, and hence is determined by what happens locally. (Hopefully this is not all nonsense.)</p> http://mathoverflow.net/questions/40713/an-elementary-lemma-of-commutative-algebra/40746#40746 Answer by Lennart Galinat for An elementary lemma of commutative algebra Lennart Galinat 2010-10-01T13:45:12Z 2010-10-01T14:08:02Z <p>If you want to avoid the use of Ext-groups, you could prove it like this (which is basically the same proof):</p> <p>Let 0→A→B→C→0 be a short exact sequence of R-modules with C finitely presented and assume it splits after localisation at every maximal ideal.</p> <p>Use the natural isomorphism Hom$_{R_m}(C_m,A_m)$=R$_m$⊗Hom$_R$(C,A) (which uses the flatness of localisation and the finitely presentedness of C) and the assumption to see that the map Hom$_R$(C,B)→Hom$_R$(C,C) is surjective since all of its localisations are.</p> <p>Since the giving a splitting is equivalent to this map being surjective we are done.</p> <p>P.S.: Of course one could prove "if and only if" in the statement like this.</p> <p>Counterexample in the general case: Take R=$\prod_{\mathbb N} \mathbb F_2$, I=$\sum_{\mathbb N} \mathbb F_2$ and C the cokernel of the inclusion.</p> <p>Now I claim three things: 1.) R has dimension zero</p> <p>Proof: Every element of R is an idempotent and every prime ideal in R has to contain exactly one of e or 1-e for every idempotent e in R. If we had a chain of prime ideals, the larger one would nescessarily have to contain e and 1-e for one such idempotent and so couldn't exist.</p> <p>2.) The localisation of R at every maximal ideal is a field.</p> <p>Proof: It is a zero-dimensional local ring by 1.) and reduced since R doesn't contain nilpotent elements. Thus this localisation is a field.</p> <p>3.) I is not a direct summand of R.</p> <p>Proof: Direct Summands correspond to idempotents in R. Since every element in R is idempotent we need to analyze all principal ideals. If an element has only finitely many non-zero entries the ideal created by it has only finitely many elements, thus can't be I. If on the other hand it contains infinitely many non-zero entries, the ideal created by it has uncountably many elements. Since I contains countably many elements, it can't be a direct summand of R.</p> <p>This establishes the counterexample, since the short exact sequence 0→I→R→C→0 splits in every localisation at a maximal ideal.</p>